Question

$2^{202} +202$ 除以 $2^{101}+2^{51}+1$ 的余数是多少？

What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1?$

$100$

$101$

$200$

$201$

$202$

答案：D

知识点：平方差模运算

难度评级：1880

小提示：

尝试用平方差来分解分子。

Try to factor the numerator using a difference of squares

大提示：

把 $2^{202}+202$ 写成除数的倍数加上一个小余数。

Write $2^{202}+202$ as a multiple of the divisor plus a small remainder

视频讲解：

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文字解答：

令 $m=2^{101}+2^{51}+1$ 。围绕除数分解： $\begin{aligned} &2^{202}+202 \\ &\quad =(2^{101}+1)^2-(2^{51})^2+201\text{。} \end{aligned}$ 利用平方差， $\begin{aligned} &(2^{101}+1)^2-(2^{51})^2 \\ &\quad =(2^{101}+2^{51}+1) \\ &\quad {}\cdot(2^{101}-2^{51}+1)\text{，} \end{aligned}$ 这是 $m$ 的倍数。因此 $2^{202}+202\equiv 201\pmod m\text{。}$

所以正确答案是 D。

Let $m=2^{101}+2^{51}+1.$ We factor the numerator around this divisor: $\begin{aligned} &2^{202}+202 \\ &\quad =(2^{101}+1)^2-(2^{51})^2+201. \end{aligned}$ By the difference of squares, $\begin{aligned} &(2^{101}+1)^2-(2^{51})^2 \\ &\quad =(2^{101}+2^{51}+1) \\ &\quad {}\cdot(2^{101}-2^{51}+1), \end{aligned}$ which is a multiple of $m.$ Therefore $2^{202}+202\equiv 201\pmod m.$

Thus, the correct answer is D .

2020 AMC 10B 第 22 题